ENGR4300 Test 3A and 3B Fall 2003

Transcription

1 Question 1 -- Astable Multivibrator R1 8 X U3 R2 TOPEN = CC TRIGGER RESETOUTPUT CONTROL THRESHOLD DISCHARGE GND 555D R3 1Meg C1 C2 10uF.01uF The circuit above has been simulated using PSpice. Using PROBE, the voltages at pins 2, 6, 7, and 3 have been displayed. a. Label which trace goes with which pin (2,3,6,7) in each time period. Be sure that you label the traces in both the on and off parts of the pulse cycle. (8 points) b. What is the duty cycle of the pulses in the plot? (4 points) T1 = 0.96s T2=0.60s T=1.56s duty cycle = T1/T =.615 duty cycle = 61.5% (answers may vary)

2 b. Determine the values of R1 and R2 from the information in this plot. (4 points) T2 = 0.693(R2)(C1) 0.6=0.693(R2)(10EE-6) R2=0.0866EE6 R2=86.6K ohms T1=0.693(R1+R2)(C1) 0.96=0.693(R1+86.6K)(10EE-6) R1+86.6K=138.5K R1=51.9K ohms c. Test A:What could you do to increase the duty cycle of the pulses? (4 points) c. Test B:What could you do to decrease the duty cycle of the pulses? (4 points) duty cycle = T1/T = [0.693(R1+R2)C1]=[0.693(R1+2R2)C1]=[R1+R2]/[R1+2R2] R1 + 1 duty cycle = R2 If R1>>R2 then the duty cycle approaches 100% -- It R1 + 2 R2 increases. If R1<<R2 then the duty cycle approaches 50% -- It decreases. Changing the value of the capacitor will influence the frequency, but not the duty cycle. Test A: increase R1 or decrease R2 Test B: increase R2 or decrease R1

3 Question 2 Measuring Inductance (20 points) This page is Test A: Given the following circuit and its AC sweep below: a) Find the transfer function, H(jω), at point B. Determine the function and the magnitude and the phase at high and low frequencies. (8 points) (2 pts) H(jω) = method 1: excluding R2 H(jω)=[R1+jωL1]/[R1+jωL1+1/jωC1] H(jω)=[jωR1C1-ω 2 L1C1]/[jωR1C1--ω 2 L1C1+1] method 2: including R2 H(jω)=[R1+jωL1]/[R2+R1+jωL1+1/jωC1] H(jω)=[jωR1C1-ω 2 L1C1]/[jω(R1+R2)C1--ω 2 L1C1+1] either is ok (1 pt) H(jω lo ) =jωr1c1 (1 pt) H(jω hi ) = 1 (1 pt) H(jω lo ) = 0 (1 pt) H(jω hi ) = 1 (1 pt) H(jω lo ) = π/2 (1 pt) H(jω hi ) = 0

4 Question 2 Measuring Inductance (20 points) This page is Test B: Given the following circuit and its AC sweep below: a) Find the transfer function, H(jω), at point B. Determine the function and the magnitude and the phase at high and low frequencies. (8 points) (2 pts) H(jω) = method 1: excluding R2 H(jω)=[1/jωC1]/[R1+jωL1+1/jωC1] H(jω)=[1]/[jωR1C1--ω 2 L1C1+1] method 2: including R2 H(jω)=[jωC1]/[R2+R1+jωL1+1/jωC1] H(jω)=[1]/[jω(R1+R2)C1--ω 2 L1C1+1] either is ok (1 pt) H(jω lo ) = 1 (1 pt) H(jω hi ) = 1/-ω 2 L1C1 (1 pt) H(jω lo ) = 1 (1 pt) H(jω hi ) = 0 (1 pt) H(jω lo ) = 0 (1 pt) H(jω hi ) = π

5 b) Based on your results from part a), indicate on the plot the trace for the magnitude of the voltage at point B (2 points). c) Find the resonant frequency f 0 from the plot. Notice that the x-axis has logarithmic scale. (ie = 3.16) (3 points) Test A: f 0 =10EE(5.7)=501,187 f 0 =500K Hz (answers may vary) Test B: f 0 =10EE(5.4)=251,189 f 0 =250K Hz (answers may vary) d) Solve for the unknown inductance. (5 points) Test A: f 0 =1/[2πsqrt(L1C1)] L1=1/[C1(2πf 0 ) 2 ] L1=1/(0.01EE-6)(πEE+6) 2 =1/[0.0987EE+6]=10.1EE-6 L1 = 10 uf (answers may vary) Test B: f 0 =1/2πsqrt(L1C1) L1=1/C1(2πf 0 ) 2 ] L1=1/(0.022EE-6)(5πEE+5) 2 =1/[5.43EE+4]=0.18EE-4 L1 = 18 uf (answers may vary) e) Both of the resistances in this circuit are not physical resistors. They represent the resistance of something else in the circuit. Assuming that these do not represent wire resistance, what do they represent? (2 points) The resistor R2 is the 50 ohm impedance of the function generator. The resistor R1 is the resistance in the wire of the inductor, L1.

6 Question 3 -- Op Amp Differentiator (20 points) Test A: Given: C1=0.001 µf R2= 500 Ω Test B: Given: C1= µf R2= 1000 Ω In this question, we will look at the effect of the 50 ohm input resistance on the opamp differentiator. a) What is the expression for out/in for circuit A above. Give this in terms of C1, R2, j and ω. (3 points) out in = jω R2C1 b) Determine the expression for out/in for circuit B above. (Be sure to include the impedance of the function generator, R1.) Give this in terms of C1, R1, R2, j and ω. (3 points) out in jω R2C1 = 1+ jω R1C 1 c) Find the corner frequency f c for circuit B. This should be a numerical value in Hertz. (4 points) Test A: f c =1/[2πR1C1]=1/[2π(50)(0.001EE-6)]=3.183EE+6 f c =3,183K Hz = 3.2Meg Hz Test B: f c =1/[2πR1C1]=1/[2π(50)(0.0022EE-6)]=1.447EE+6 f c =1,447K Hz = 1.4Meg Hz

7 d) Does circuit B behave like circuit A above or below the corner frequency? Use limits to show how you know this. (6 points) out in jω R2C1 = 1+ jω R1C 1 at low frequencies : at high frequencies : out in out in jω R2C1 = = jω R2C1 differentiator( likea) 1 jω R2C1 R2 = = ( inverting amplifier) jω R1C 1 R1 Circuit B will behave like circuit A below the corner frequency. e) At about what frequencies does the 50 ohm impedance of the function generator prevent this circuit from differentiating? (4 points) Circuit B should act like a differentiator at frequencies much less than f c. Therefore, it will not differentiate above f c. Minimally we could say that the resistance prevents this circuit from working at frequencies above about 3 Meg Hz for Test A and 1.5 Meg Hz for Test B. Even better, we could estimate that it would not be differentiating even above frequencies a bit lower, perhaps 1Meg Hz for Test A and 1.5Meg Hz for Test B. For a conservative estimate, we have plenty of frequency range available to go down an entire decade. Therefore, I would say conservatively, Test A: 300K Hz and above Test B: 150K Hz and above (answers may vary)

8 Question 4 Zener Diodes (20 points) The circuit above is a zener diode voltage regulator. Assume the zener voltage of the diode is 4.7 volts and its forward bias voltage is 0.7 volts. 1) What does a voltage regulator do? (2 points) A voltage regulator is a circuit which prevents the voltage from exceeding a set voltage limit. (In this case, the positive limit is +4.7 and the negative limit is 0.7.) 2) What would the voltage at 2 be when 1 = 6 if R2 has the following values: (6 points) i) 1K ohms expected voltage 2=(1K/11K)6=0.54 (0.54<4.7) 2=0.54 ii) 10K ohms expected voltage 2=(10K/20K)6=3 (3<4.7) 2=3 iii) 100K ohms expected voltage 2=(100K/110K)6=5.45 (5.45>4.7) 2=4.7 3) What would the voltage at 2 be when 1 = -6 and if R2 has the following values: (6 points) i) 1K ohms expected voltage 2=(1K/11K)-6=-0.54 (-0.54>-0.7) 2 = ii) 10K ohms expected voltage 2=(10K/20K)-6=-3 (-3<-0.7) 2 = -0.7 iii) 100K ohms expected voltage 2=(100K/110K)-6=-5.45 (-5.54<-0.7) 2 = -0.7

9 3) Each of the following three plots was made with a different value for R2: 1K ohms, 10K ohms and 100K ohms. Indicate which plot was made with which value for R2. (6 points) The plot below cuts off at 4.7 and > R2=100K The plot below doesn t cut off at all --> R2=1K The plot below cuts off at 0.7, but not at > R2=10K

10 Question 5 -- Circuit functionality and transformers (20 points) The following circuit was constructed to test two DC power supplies. One is a battery and one is a wall wart (connects to a normal 120 outlet). The boxes surrounding each part of the circuit identify the functional blocks (each has a specific purpose). Each of the boxes is also shown expanded for clarity. A Rs R3 B TXs R4 D2 OFF = 0 AMPL = 120 FREQ = 60 s D1N4148 C2 33uF 0 C D E 2 OFF = 0 AMPL = 100m FREQ = 1kHz R7 50 C3 F.1uF 3 11 G R5 10k U ua OS2 5 OUT 6 OS1 1 R6 100k H Rl 1Meg I A Rs R3 B TXs R4 D2 OFF = 0 AMPL = 120 FREQ = 60 s D1N C E R7 C2 33uF D 2 OFF = 0 AMPL = 100m FREQ = 1kHz 50 C3.1uF F

11 3 11 G R5 10k U ua OS2 OUT OS1 R6 100k H Rl 1Meg I a. Identify the function of each of the nine blocks. (Draw a line to connect the letter of the block to its function). (9 points) b. On the next page are plotted six voltages measured at various points in the circuit. Identify each of the voltages by indicating the block for which this is the output voltage. Note that there are only six voltages but there are eight output points for the blocks. (6 points) c. Based on the voltages you have just identified, what is the ratio of the input voltage to the output voltage of the transformer? Note that a real transformer is modeled here so that it has finite resistance in its windings. However, you can neglect these small resistances in the rest of this problem. (3 points) 120 : 12 = 10 : 1 d. If the primary winding of the transformer has turns, how many turns does the secondary winding have to produce this change in voltage? (2 points) Test A: N2/10000 = 1/10 N2 = 1000 turns Test B: N2/20000 = 1/10 N2 = 2000 turns

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